Question

Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = 4x2 + 5x – 1.

Answer 1

the axis of symmetry for a quadratic function is x=-b/2a
in this case, b=5, a=4, so the line of symmetry is x=-5/8
find out the value of y when x=-5/8: y=4(-5/8)² +5(-5/8)-1=-41/8
so the vertex is at (-5/8, -41/8)
double check my calculation please.

Answer 2

Vertex:

`(-0.625,-2.563)`

Axis of symmetry:

`x=-0.625`

Explanation:

The function is given by the equation:

`y=4x^2+5x-1`

In general for any quadratic equation of the type:

`y=ax^2+bx+c`

The coordinate of the vertex is: (h,k)

where

`h=(-b)/(2a)`

and

`k=(-(b^2-4ac))/(4a)`

and the equation of axis of symmetry is:

`x=h`

Here we have:

`a=4,\ b=5\ and\ c=-1`

i.e.

`h=(-(5))/(2* 4)\\\\h=(-5)/(8)=-0.625`

and

`k=(-(5^2-4* 4* (-1))/(4* 4)\\\\\\k=(-(25+16))/(16)\\\\k=(-41)/(16)=-2.563`

Hence, the vertex is:

`(-0.625,-2.563)`

and the axis of symmetry is given by:

`x=-0.625`